Quadrilateral's area problem

Question

I have some troubles with this problem :

Let $ABCD$ be a convex quadrilateral. $M$, $N$, $P$ and $Q$ are the midpoints of the sides $AB$, $BC$, $CD$ and $AD$. $AN$, $BP$, $MD$ and $CQ$ are interescting in $X$, $Y$, $Z$ and $T$ like in the figure below. Prove that $[XYZT] = [AMX] + [BYN] + [CZP] + [DTQ]$. It is noted with $[ABC]$ the area of $ΔABC$.

Since $M$, $N$, $P$ and $Q$ are midpoints, the first thing that came in my mind was the median property : the median divides a triangle in two echivalent triangles (with the same area).

I would appreciate some suggestions.

Thanks!

Answer 1

Let $$[AMX]=a,[XMBY]=b,[BYN]=c$$ $$[AXTQ]=d,[XYZT]=e,[YNCZ]=f$$ $$[QTD]=g,[TZPD]=h,[ZCP]=i.$$

From $$[BCP]=[BPD]\quad \text{and}\quad[AMD]=[BMD],$$

$$[ABCD]=2(c+f+i)+2(a+d+g)\tag1$$

From $$[ABN]=[ANC]\quad\text{and}\quad[ACQ]=[QCD],$$

$$[ABCD]=2(a+b+c)+2(g+h+i)\tag2$$ From $(1)(2)$, $$2(c+f+i)+2(a+d+g)=2(a+b+c)+2(g+h+i)\Rightarrow f+d-b-h=0\tag3$$

From $(1)(3)$, we have $$a+b+c+d+e+f+g+h+i=2(c+f+i)+2(a+d+g)$$ $$\Rightarrow e=a+c+i+g+(f+d-b-h)=a+c+i+g,$$ i.e. $$[XYZT]= [AMX] + [BYN] + [CZP] + [DTQ].$$

Answer 2

First, one can notice that $2[ABN]=[ABC]$, since $N$ is the middle of $BC$. From that (applied to the four corner triangles), we deduce that $[ABN]+[BPC]+[CQD]+[DAM]=[ABCD]$. Now one can also observe that $[ABN]+[BPC]+[CQD]+[DAM]=[ABCD]-[XYZT]+([AMX] + [BYN] + [CZP] + [DTQ])$. By combining those equations we get the desired result.